Partial Slopes and Predictor Units

Introduction

A regression slope is not a unit-free description of a relationship. A slope of 2 may mean two dollars per policy, two thousand dollars per policy, or two dollars per thousand policies, depending on how the response and predictor are measured. In multiple regression, it also describes a partial change: the fitted response change associated with changing one predictor while holding the others fixed.

Changing a predictor’s measurement unit changes the numerical value of its coefficient. It does not, when the conversion is applied consistently, change the fitted responses. That distinction matters when interpreting a model, comparing output produced in different units, or checking a calculation. This topic is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Definition: In a multiple linear regression model with fitted equation $\hat{Y}=b_0+\sum_{j=1}^{p}b_jx_j$, the coefficient $b_j$ is the fitted response change for a one-unit increase in $x_j$, holding every other predictor fixed. Its units are response units per predictor unit. For a change of $\Delta x_j$ units, with all other predictors unchanged, the fitted response changes by $b_j\Delta x_j$.

Partial Slopes Have Units

The model underlying the fitted equation is

Y_i=\beta_0+\sum_{j=1}^{p}\beta_jx_{ij}+\varepsilon_i.

Here $Y_i$ is the response for observation $i$, $x_{ij}$ is its value of predictor $j$, and $\varepsilon_i$ is the error. For the fitted equation, $b_0,b_1,\ldots,b_p$ are the estimated coefficients. When interpreting the equation as a conditional mean, the usual mean-zero condition is $E[\varepsilon_i\mid x_{i1},\ldots,x_{ip}]=0$. The algebraic statements about fitted values below do not require normally distributed errors.

For a particular predictor $x_j$, keep every other predictor fixed and increase $x_j$ by $\Delta x_j$. The terms involving the other predictors and the intercept cancel when the two fitted responses are subtracted:

\hat{Y}_{\text{new}}-\hat{Y}_{\text{old}} = \left(b_0+\sum_{k\ne j}b_kx_k+b_j(x_j+\Delta x_j)\right) - \left(b_0+\sum_{k\ne j}b_kx_k+b_jx_j\right) =b_j\Delta x_j.

This is why a partial slope gives a response change, not a complete description of a real-world intervention. The other predictors are held fixed in the comparison. If they change too, their terms also contribute to the fitted response difference.

Worked Example: Translating a Partial Slope Into Dollars

A hypothetical insurer models annual claim cost $Y$, measured in thousands of dollars, using policy duration $x$, measured in years, and a risk score $z$, measured in points. Its fitted equation is $\hat{Y}=18+0.75x-1.2z$. Interpret the coefficient of $x$, then compare the fitted responses at $x=6$ and $x=10$ when $z=5$.

Method. The coefficient of $x$ is $0.75$ thousand dollars per year, conditional on a fixed risk score. The predictor change is $\Delta x=10-6=4$ years, so the fitted response change is

0.75(4)=3 \text{ thousand dollars}=\$3{,}000.

Check by evaluating both fitted responses. At $x=6,z=5$,

\hat{Y}=18+0.75(6)-1.2(5)=18+4.5-6=16.5

thousand dollars. At $x=10,z=5$,

\hat{Y}=18+0.75(10)-1.2(5)=18+7.5-6=19.5

thousand dollars. The difference is $19.5-16.5=3$ thousand dollars, agreeing with the slope calculation. The common mistake this avoids is reporting $0.75$ as dollars per year: the response was measured in thousands of dollars.

How Changing Predictor Units Changes Coefficients

Suppose the model is fitted using predictor $x_j$, and a new measurement is defined by $x_j^*=a_j+c_jx_j$, where $c_j\ne 0$. The star indicates the re-expressed measurement; it does not indicate a different underlying characteristic. Solving for the original measurement gives $x_j=(x_j^*-a_j)/c_j$. Substitution into the fitted equation gives

\hat{Y} = \left(b_0-\sum_{j=1}^{p}\frac{b_ja_j}{c_j}\right) +\sum_{j=1}^{p}\frac{b_j}{c_j}x_j^*.

Thus the new slope for predictor $j$ is $b_j/c_j$, and the intercept becomes $b_0-\sum_j b_ja_j/c_j$. If the conversion only multiplies the predictor by $c_j$, then $a_j=0$, so the intercept is unchanged. If the conversion also shifts the origin, the intercept generally changes as well.

Key result: A consistent, one-to-one change of predictor units changes coefficient values but leaves the fitted response at each observation unchanged. Scaling $x_j^*=c_jx_j$ divides its coefficient by $c_j$. Shifting and scaling $x_j^*=a_j+c_jx_j$ also adjusts the intercept to account for the changed origin.

The result follows directly from substitution, but it also has an optimization explanation. Ordinary least squares chooses coefficients to minimize the residual sum of squares. Because $x_j$ and $x_j^*=a_j+c_jx_j$ determine one another when $c_j\ne0$, each coefficient choice in the original model has a corresponding coefficient choice in the re-expressed model that produces exactly the same fitted responses. The residuals, and therefore the residual sum of squares, are identical for each corresponding choice. Consequently, the least-squares fit is unchanged. If the original design matrix has full column rank, the coefficient vector is unique; the transformed coefficients are the corresponding unique re-expression of that fit.

Worked Example: Converting Celsius to Fahrenheit in a Multiple Regression

A hypothetical energy model has fitted equation $\hat{Y}=40+2C+3z$, where $C$ is temperature in degrees Celsius and $z$ is another predictor. Re-express the equation using Fahrenheit $F=32+1.8C$, then find the fitted response at $F=50,z=2$ and verify it in the original units.

Method. First solve the unit conversion for $C$:

C=\frac{F-32}{1.8}=\frac{5}{9}(F-32)=\frac{5}{9}F-\frac{160}{9}.

Substitute this expression into the fitted equation:

\begin{aligned} \hat{Y} &=40+2\left(\frac{5}{9}F-\frac{160}{9}\right)+3z\\ &=40+\frac{10}{9}F-\frac{320}{9}+3z\\ &=\frac{40}{9}+\frac{10}{9}F+3z. \end{aligned}

At $F=50,z=2$, the re-expressed equation gives

\hat{Y} =\frac{40}{9}+\frac{10}{9}(50)+3(2) =\frac{40}{9}+\frac{500}{9}+6 =60+6=66.

Verify using the original predictor units. Since $C=(50-32)/1.8=18/1.8=10$, the original equation gives

\hat{Y}=40+2(10)+3(2)=40+20+6=66.

Both expressions give $66$. The common mistake is to use the conversion in the wrong direction or to calculate the transformed intercept incorrectly. The new slope is $2/1.8=10/9$, and the intercept includes the shift from $32$, so it is $40/9$, not $40$. No rounding is needed.

Worked Example: Changing From Hundreds to Thousands of Policies

A hypothetical model predicts a response $Y$ in dollars from a claim-rate predictor $x$, measured as claims per 100 policies, and another predictor $z$. The fitted equation is $\hat{Y}=120+24x+5z$. Re-express the claim rate as $x^*$, measured as claims per 1,000 policies, and compare the fitted response at $x=3,z=4$ with the transformed equation.

Method. A rate of $x$ claims per 100 policies corresponds to $10x$ claims per 1,000 policies, so $x^*=10x$, or $x=x^*/10$. Substitute:

\hat{Y}=120+24\left(\frac{x^*}{10}\right)+5z =120+2.4x^*+5z.

The transformed slope is $24/10=2.4$ dollars per one-unit increase in the rate measured per 1,000 policies. The other slope and intercept stay unchanged because neither the response units nor the other predictor’s units changed.

At $x=3,z=4$, the new rate is $x^*=10(3)=30$. In the original equation,

\hat{Y}=120+24(3)+5(4)=120+72+20=212.

In the transformed equation,

\hat{Y}=120+2.4(30)+5(4)=120+72+20=212.

The fitted responses agree. As an additional check, an increase of $15$ units in $x^*$ corresponds to an increase of $15/10=1.5$ units in $x$. The response change is $2.4(15)=36$ dollars using the transformed slope and $24(1.5)=36$ dollars using the original slope. The common mistake is multiplying the coefficient by 10 because the new predictor values are ten times larger; the coefficient must instead be divided by 10.

Worked Example: Scaling One Predictor and Shifting Another

A hypothetical fitted equation is $\hat{Y}=7+4x-2w+0.5z$. A reporting system defines $x^*=10x$ and $w^*=w-5$, leaving $z$ unchanged. Rewrite the equation and verify the result at $x=3,w=8,z=2$.

Method. The inverse conversions are $x=x^*/10$ and $w=w^*+5$. Substituting both conversions gives

\begin{aligned} \hat{Y} &=7+4\left(\frac{x^*}{10}\right)-2(w^*+5)+0.5z\\ &=7+0.4x^*-2w^*-10+0.5z\\ &=-3+0.4x^*-2w^*+0.5z. \end{aligned}

Here the coefficient of $x^*$ is $4/10=0.4$. Since $w^*=w-5$, the change in origin for $w$ contributes $-2(5)=-10$ to the intercept, changing it from $7$ to $-3$.

For the stated values, $x^*=10(3)=30$ and $w^*=8-5=3$. The original equation gives

\hat{Y}=7+4(3)-2(8)+0.5(2)=7+12-16+1=4.

The transformed equation gives

\hat{Y}=-3+0.4(30)-2(3)+0.5(2)=-3+12-6+1=4.

The match checks both the scaling and the shifted intercept. The common mistake is to update the scaled slope but ignore the intercept effect of shifting $w$.

What Changes, and What Does Not

A coefficient’s numerical value depends on the measurement units, but its interpretation follows the units in the equation. A slope can be translated into the response change for any specified predictor increment by multiplying by that increment. The resulting response change does not depend on whether the predictor is recorded in one unit or a rescaled unit.

An intercept is the fitted response when every predictor equals zero in the units used in the model. A shift in a predictor’s origin changes which physical value is represented by zero and can therefore change the intercept. The intercept need not have a useful practical interpretation if that all-zero predictor combination is outside the relevant range.

A change in predictor units is different from a change in response units. If the response is also converted, the entire fitted equation, including the intercept and all slopes, must be converted into the new response units. In this tutorial’s transformations, the response units are held fixed, so only coefficients affected by predictor conversions change.

Common Mistakes

Ignoring the other predictors. A partial slope describes a change in one predictor while holding the other predictors fixed. If another predictor changes, include its coefficient times its change too.
Forgetting the response units. A coefficient measured in thousands of dollars per year does not represent dollars per year until it is multiplied by 1,000.
Applying the conversion in the wrong direction. If $x^*=10x$, then $x=x^*/10$, so the coefficient becomes $b/10$, not $10b$.
Leaving the intercept unchanged after shifting a predictor. For $x^*=a+cx$, use $x=(x^*-a)/c$ in the full equation. The offset can change the intercept.
Changing just one part of the equation. Convert predictor values and coefficients consistently. Unchanged predictors keep their coefficients, and the fitted response should agree under either representation.

Key takeaway: A partial slope is measured in response units per predictor unit. To interpret a predictor increment, multiply its slope by that increment while holding the other predictors fixed. Under a consistent change of predictor units, coefficients may change, but the fitted responses do not.

Check Your Understanding

For each problem, track the units and use the stated conversion consistently.

A fitted equation is $\hat{Y}=9+1.4u-0.5v$, where $Y$ is measured in thousands of dollars. With $v$ fixed, how much does the fitted response change when $u$ increases by $2.5$? State the change in dollars. (Answer: $1.4(2.5)=3.5$ thousand dollars, or $\$3{,}500$.)
A fitted equation is $\hat{Y}=6+0.03x+2z$, where $x$ is measured in dollars. Define $x^*=x/100$, so $x^*$ is measured in hundreds of dollars. Rewrite the equation and verify its fitted response at $x=500,z=1$. (Answer: $\hat{Y}=6+3x^*+2z$; at $x^*=5,z=1$, $\hat{Y}=6+15+2=23$, matching the original equation.)
A model is $\hat{Y}=11+3C-4z$, and a temperature conversion is $R=50+2C$. Rewrite the model using $R$, then find the fitted response at $R=70,z=2$. (Answer: $C=(R-50)/2$, so $\hat{Y}=1.5R-64-4z$; at $R=70,z=2$, $\hat{Y}=105-64-8=33$.)
A model is $\hat{Y}=30+5x-6v+2z$. Define $x^*=x+4$ and $v^*=2v$. Rewrite the equation and check it at $x=1,v=3,z=2$. (Answer: $x=x^*-4$ and $v=v^*/2$, so $\hat{Y}=10+5x^*-3v^*+2z$; at $x^*=5,v^*=6,z=2$, the fitted response is $10+25-18+4=21$, matching the original.)
A predictor is recorded as $q$ units per 50 exposures, and its fitted slope is $8$ response units per one-unit increase in $q$. A new system records the same rate per 500 exposures, so $q^*=10q$. With other predictors fixed, what is the new slope, and what response change corresponds to an increase of $25$ units in $q^*$? (Answer: The new slope is $8/10=0.8$; the response change is $0.8(25)=20$ response units, equivalent to an increase of $2.5$ in $q$ and a change of $8(2.5)=20$.)

What You'll Learn